Download Trigonometry

CAMOSUN COLLEGE
School of Access
Academic and Career Foundations Department
MATH 052 unit 5
Trigonometry
adapted from:
ABE Intermediate Level Mathematics
Module 14:
Trigonometry
1
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ISBN 0-7718-9544-5 (set)
1. Mathematics – Study and teaching (Continuing education) – British Columbia.
I. Brodie, Sarah. II. British Columbia. Ministry of Advanced Education, Training and Technology.
III. Centre for Curriculum, Transfer and Technology. IV. Series.
QA14.C32B742 1999
510’.71’5
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ABE Intermediate Level Mathematics
Module 1: Arithmetic and Estimation
Module 2: Measurement
Module 3: Perimeter, Area and Volume
Module 4: Ratio and Proportion
Module 5: Percent
Module 6: Geometry: Introduction
Module 7: Geometry: Construction
Module 8: Statistics
Module 9: Signed (Rational) Numbers
Module 10: Algebra: Equations
Module 11: Algebra: Polynomials
Module 12: Powers, Roots and Scientific
Notation
Module 13: Graphing
Module 14: Trigonometry
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Copyright  1999 Province of British Columbia
Ministry of Advanced Education, Training and Technology
This document may not be reproduced in any form without permission of the
Centre for Curriculum, Transfer and Technology.
2
Adult Basic Education
Intermediate Level Mathematics
Module 14
Trigonometry
Prepared by
Sarah Brodie, North Island College
Pat Corbett, North Island College
Paul Grinder, Okanagan University College
Peter Robbins, Kwantlen University College
Ada Sarsiat, Northwest Community College
for the
Province of British Columbia
Ministry of Advanced Education, Training and Technology
and the
Centre for Curriculum, Transfer and Technology
1999
Revised 2000
3
Contents
Learning outcomes .................................................................... 5
Introduction ............................................................................... 6
5.1 The right triangle ................................................................ 7
Exercise 5.1 ..................................................................... 8
5.2 Angles and sides................................................................. 9
Exercise 5.2 ................................................................... 10
5.3 Pythagorean Theorem ...................................................... 12
Exercise 5.3 ................................................................... 14
5.4 The tangent ratio .............................................................. 17
Exercise 5.4 ................................................................... 19
5.5 Using the tangent ratio ..................................................... 21
Exercise 5.5 ................................................................... 23
5.6 The sine and cosine ratios ................................................ 26
Exercise 5.6 ................................................................... 28
5.7 Solving triangles .............................................................. 32
Exercise 5.7 ................................................................... 34
Practice Test ............................................................................ 36
Problem Set A ......................................................................... 38
Problem Set B ......................................................................... 40
Appendix A: Table of Trigonometric Ratios ......................... 41
Appendix B: Glossary ............................................................ 42
Answers................................................................................... 43
4
Learning outcomes
Trigonometry is an application of ratio and proportion, geometry, algebra, and measurement.
Though it is usually considered an advanced math topic, it is applied in various levels of trades and
technology.
When you have completed this module, you should be able to:
name the parts of a right triangle
find the missing side of a right triangle using the Pythagorean Theorem
find the measure of an unknown side or angle of a right triangle using sine, cosine or tangent
ratios
solve problems using right triangle trigonometry




Procedure for independent study
You will require a scientific calculator, ruler, and protractor.
1.
Study each of the units in order, complete all of the exercises, and check your answers.
If you need assistance, contact your instructor.
2.
Review the definitions and terminology in the Glossary (p 42).
3.
For more practice on application problems, complete Problem Sets A & B (p 38−40).
4.
Complete the Practice Test (p 36) and check your answers.
5.
Complete the MATH 052 unit 5 test.
5
Introduction
Trigonometry means “triangle measure”. It is a branch of mathematics which was originally
developed to calculate unknown sides and angles in triangles. Trigonometry is very useful in
applications such as surveying, navigation, and electronics.
Example 1
Suppose you wanted to know the height of a tree without having to climb the tree with a measuring
tape. You can find the height by taking two measurements from the ground (Figure 1).
Figure 1
Once the distance from the base of the tree is known, as well as the
angular measurement to the top of the tree, you can make a scale
drawing of a similar triangle and measure the length of the
corresponding side (Figure 2).
Using the proportion:
x
35.8 cm
=
30 m
30 cm
Figure 2
the tree is 35.8 m tall.
By using trigonometry, you can solve problems like this in an even simpler fashion.
6
5.1
The right triangle
The right triangle in Figure 3 is labelled using standard lettering.
The three angles of the triangle are A, B and C. The right
angle is C = 90.
The sides of the triangle are a, b and c. The line segments AB ,
AC and BC also denote the sides.
Side c is called the hypotenuse of the triangle while sides a and b
are called the legs of the triangle. The hypotenuse is always the
longest side of the triangle.
The Pythagorean Theorem states that for any right triangle,
c2 = a2 + b2
Figure 3
Angles of triangles are designated with capital letters. Sides are designated with lower case letters.
In a right triangle (Figure 4), sides are related to angles as follows:



side a is opposite A
side b is opposite B
side c is opposite C
In a right triangle, the hypotenuse is always opposite the
right angle. Note that the names of angles A and B
could be reversed and the same for sides a and b. Only
side c and angle C are fixed.
Notice that



the side opposite an angle is not part of the angle
side a is adjacent to B
side b is adjacent to A
Figure 4
Sides that are adjacent to angles are actually part of the angle. The hypotenuse is already named
and is not considered to be an adjacent side of an angle. Also, we do not consider the right angle to
have adjacent sides.
Now complete Exercise 5.1 and check your answers. You may have to sketch
triangles and take measurements with a ruler and protractor to answer some
questions.
7
Exercise 5.1
1.
All triangles have three sides and three angles.
a. true
2.
The sum of the measures of the three angles of any triangle is equal to 180.
a. true
3.
b. false
The sides that form the right angle are perpendicular to each other.
a. true
10.
b. false
The side opposite the right angle is called the hypotenuse.
a. true
9.
b. false
All right triangles have one right angle.
a. true
8.
b. false
The area of a triangle is equal to the product of ½ the base times the height.
a. true
7.
b. false
The perimeter of a triangle is equal to the product of ½ the base times the height.
a. true
6.
b. false
The side opposite the smallest angle of a triangle has the shortest length. The side opposite
the largest angle has the longest length.
a. true
5.
b. false
The sum of the length of any two sides of a triangle is always longer than the third side.
a. true
4.
b. false
b. false
The Pythagorean Theorem states that the sum of the squares of the legs of a right triangle is
equal to the square of the hypotenuse.
a. true
b. false
Check your answers with the answer key, located at the end of this unit.
8
5.2
Angles and sides
The sum of the interior angles of any triangle is 180. In any
right triangle (Figure 5), one of the angles is 90. The sum of
the other two angles is also 90 or
A + B = 90
Figure 5
Example 1
Find B of the triangle in Figure 6.
Solution
Using the formula:
A + B = 90
28 + B = 90
Figure 6
B = 90 − 28
B = 62
It is very important that you always know precisely which angles and sides are
involved in your calculations.
Now complete Exercise 5.2 and check your answers.
9
Exercise 5.2
1.
In the triangle shown, indicate:
B
a. which side is the hypotenuse
c
b. which side is opposite B
a
c. which side is adjacent to A
o
40
d. which side is adjacent to B
C
A
b
e. the measure of B
2.
In PQR, indicate:
R
a. the hypotenuse
p
b. the side opposite Q
Q
q
c. the side adjacent to P
r
d. the side adjacent to Q
e. the side opposite P
P
f. the side opposite R
g. if P = 20 then Q =
10
3.
Determine x in each of the following triangles.
4.
In the right triangle shown, identify the large (L), middle (M) and small (S) angles. Also
identify the long (l), middle (m) and short (s) sides. Then answer the following questions:
a. Which angle is the long side opposite?
b. Which angle is the middle side opposite?
c. Which angle is the short side opposite?
d. Are the above statements true for any triangle?
Check your answers with the answer key at the end of this unit.
11
5.3
The Pythagorean Theorem
The Pythagorean Theorem is probably the most famous theorem
in mathematics. It was known to the ancient Chinese,
Egyptians and Greeks thousands of years ago.
c
Pythagoras’ Theorem (Figure 7) states that the square of the
hypotenuse of a right triangle equals the sum of the square of its
legs or:
b
c2 = a2 + b2
a
This theorem allows us to find the length of the third side of any
right triangle when we know the length of two of the other
sides.
Figure 7
Example 1
Find the hypotenuse of the triangle in Figure 8. Round your answer to one decimal place.
Solution
Using the formula
c2 = a2 + b2
c2 = 62 + 4.52
c2 = 36 + 20.25
c2 = 56.25
c =
Figure 8
56.25
c = 7.5
12
Example 2
Find the missing side of the right triangle in Figure 9. Round your answer to one decimal place.
Solution
The missing side of the right triangle is a leg. It can be named a or b. For this example, you can
name it b.
c2 = a2 + b2
13.22 = 11.52 + b2
174.24 = 132.25 + b2
174.24 - 132.25 = b2
41.99 = b2
b =
41.99
Figure 9
b = 6.4799…
b  6.5
Note that many of the answers that we will calculate will be irrational numbers (decimal numbers
that do not terminate and do not repeat in a predictable way). These numbers are “approximate”
numbers, which can be rounded to a specified number of decimal places.
Now complete Exercise 5.3 and check your answers.
13
Exercise 5.3
1.
Name the right angle in each of the triangles shown below.
a.
C
b.
B
N
c.
O
d.
A
Z
X
B
A
C
Y
M
2.
Name the hypotenuse in each triangle shown below.
3.
Find the missing side, x, in each triangle shown below. Round your answers to one decimal
place.
14
4.
The length of a rectangle is 52 cm and the width is 40 cm. Find the length of the
diagonal. Round your answer to one decimal place.
15
5.
A 4 m ladder leans against a building. If the bottom of the ladder is 1.3 m from the bottom
of the building, how far up the wall will the ladder reach? Round your answer to one
decimal place.
6.
A warden observes a fire 4.8 km directly east of his cabin. The nearest water bomber is
stationed 10.5 km directly south of the cabin. How far is the water bomber from the fire?
Round your answer to one decimal place.
7.
A basketball player is 2.3 m tall. He has to make a phone call from a
pay phone booth which is 1 m wide, 1 m long and 2 m high. Can he fit
himself into the phone booth without sitting down or bending? (Hint:
What is the distance between opposite corners of the phone booth?)
Answer key at the end of this unit.
16
5.4
The tangent ratio
The three right triangles in Figure 10 are called similar triangles. They are similar triangles because
their corresponding angles are equal. Notice that their sides are not equal. With a metric ruler,
measure the opposite and adjacent sides (to the 35 angle) in each triangle and record your results in
the table. Round your answers to 0.1 cm.
TRIANGLE
OPPOSITE
SIDE
ADJACENT
SIDE
1
2.8 CM
4.0 CM
RATIO OF
2.8
4.0
OPPOSITE
ADJACENT
DECIMAL
EQUIVALENT
0.7
2
3
Figure 10
The table should show that the ratio of opposite side to adjacent side is the same for all three
triangles with a 35 angle, regardless of the size of the triangle. This is the core idea of
trigonometry.
17
For any acute angle of a right triangle, the ratio of the opposite side to the adjacent side (Figure 11)
is called the tangent ratio, abbreviated “tan”.
For any right triangle:
tan A or tan A =
oppositeside
adjacent side
From your investigation of the three triangles in Figure
10, you can now write:
tan 35 =
Figure 11
28
= 0.7
40
You can use a scientific calculator to determine the value of tan 35 by pressing
depending on your particular make of calculator.
35 TAN or TAN 35
Make sure your calculator is in degree mode (DEG) first.
You should get approximately 0.700208 (to nearest millionth).
Example 1
Use a metric ruler to determine tan 60 in Figure 12. Also determine tan 60 using your calculator.
Solution
Measure the opposite and adjacent sides of the triangle, then:
tan 60 =
4.6
= 1.7
2.7
The calculator will show 1.732051 for tan 60.
Again, note that tangent values to the nearest millionth in your
calculator are much more accurate than the calculated values
based on measurements to the nearest tenth.
Figure 12
Now complete Exercise 5.4 and check your answers.
18
Exercise 5.4
1.
Use a ruler, then a calculator to find the tangents of the indicated angles. Record your
results in the table provided.
cm
cm
cm
o
10
o
20o
45
cm
cm
cm
cm
cm
50o
75o
cm
cm
Angle
length of opposite side
decimal equivalent
calculator value of tan A
length of adjacent side
to six decimal places
to six decimal places
10
20
45
50
75
2.
Use a calculator to find the following. Round your answers to four decimal places.
a. tan 0
b. tan 1
c. tan 5
d. tan 15
e. tan 22
f. tan 31.4
g. tan 45
h. tan 50.92
i. tan 64
j. tan 0.05
k. tan 86
l. tan 89
19
3.
Measure the legs of the triangle to find the following.
Round your answers to one decimal place.
a. tan 25
b. tan 65
4.
Use a calculator to find the following. Round your answers to three decimal places.
a. tan 88
b. tan 89
c. tan 89.9
d. tan 89.99
e. tan 90
5.
These values are very large. When a right triangle has a
very large acute angle, the opposite side is much larger
than the adjacent side.
Find tan 90. Explain why tan 90 is undefined.
Answer key at the end of this unit.
20
5.5
Using the tangent ratio
If one acute angle and one leg of a right triangle are known, the tangent ratio can be used to find the
other leg of the triangle.
Example 1
Find side b in the triangle in Figure 13. Round your answer to one decimal place.
Solution
Side b is opposite 40 and the known side is
adjacent to 40, so we use the tangent ratio.
tan 40 =
oppositeside b

adjacent side 13
Rewrite this as a proportion and solve by crossmultiplying.
Figure 13
tan 40
b
=
1
13
1 ∙ b = 13 ∙ tan 40 or 1 ∙ b = (tan 40) ∙ 13
b  10.908  10.9
To find 13 ∙ tan 40, depending on the type of calculator,
press 13 × TAN 40 =
or
13 × 40 TAN =
Example 2
Find side a of the triangle in Figure 14. Round your
answer to one decimal place.
Solution
Side a is adjacent to 65° and the known side is
opposite 65°, so we use the tangent ratio.
Figure 14
oppositeside 24
tan 65 =

adjacent side a
21
Rewrite and solve as a proportion:
tan 65 24
=
1
a
(tan 65)  a = 1  24
a=
cross - multiply
24
 11.191
tan 65
divide by tan 65 
a  11.2
To find
24
, depending on the type of calculator,
tan 65
press 24 ÷ TAN 65 =
or
24 ÷ 65 TAN =
You can also use the tangent ratio to find an acute angle of the right triangle when both legs are
known.
Example 3
Find angle A of the right triangle in Figure 15. Round your answer
to the nearest degree.
Solution
The known sides are opposite and adjacent to angle A,
so we use the tangent ratio.
tan A =
oppositeside 16
  1.230769
adjacent side 13
To find angle A, with the calculator displaying its tangent of
1.230769, press INV TAN or 2ndF TAN or SHIFT TAN or
TAN−1 , to display the angle 50.906° or 50.9°.
Figure 15
When the tangent of an angle is known, the angle is called the “inverse tangent” or “arc tangent”.
Now complete Exercise 5.5 and check your answers.
22
Exercise 5.5
1.
Find side x in each of the following triangles. Round your answer to one decimal place.
oppositeside
Remember: tan A =
.
adjacent side
23
2.
a. In ABC, C = 90, B = 55 and AC = 18.
Sketch and label ABC, then find BC .
b. In ABC, C = 90, A = 12 and AC = 43.
Sketch and label ABC, then find BC .
3.
Find angle x in each of the following triangles. Round answers to one decimal place.
oppositeside
Remember: tan x =
.
adjacent side
4. In ABC, C = 90, AC = 5.3 and BC = 7.5.
Sketch and label ABC, then find A and B.
24
5.
Find the missing angles and sides in ABC.
Hints: Use A + B = 90 to find B.
Use the tangent ratio to find AC .
Use the Pythagorean Theorem to find AB .
6.
Solve ABC. (find A, AC and AB ).
7.
A road rises 6 m vertically for every 50 m of
horizontal distance. At what angle is the road
inclined?
8.
Joan stands 20 m from the
base of a large tree. Her line
of sight to the top of the tree
makes an angle of 54 with
the horizontal. Joan is 1.6 m
tall. How tall is the tree?
9.
A roof rises 1.2 m for every 2 m run. Find the pitch of the roof in degrees (angle x).
Answer key at the end of this unit.
25
5.6
The sine and cosine ratios
The tangent ratio can be used to find missing sides of right triangles. If the hypotenuse of a right
triangle is unknown, there are two other trigonometric ratios that are useful:


the sine ratio, abbreviated sin, pronounced “sign”
the cosine ratio, abbreviated cos, pronounced “kose”
For any right triangle (Figure 16):
sin A or sin A =
oppositeside
hypotenuse
cos A or cos A =
adjacent side
hypotenuse
Figure 16
Example 1
Find side AB in ABC (Figure 17).
Solution
An angle and its opposite side are known. To find the hypotenuse, use the sine ratio.
sin 41 =
8
AB
Rewrite and solve as a proportion:
sin 41
8
=
1
AB
(sin 41) AB = 8
AB =
To find
Figure 17
8
 12.194  12.2
sin 41
8
on a calculator,
sin 41
press 8 ÷ SIN 41 =
or
8 ÷ 41 SIN =
26
Example 2
Find side x of the triangle in Figure 18.
Solution
Side x is adjacent to the 68 angle and the hypotenuse is
9.6 cm. The cosine ratio can be used to find side x.
Figure 18
x
9.6
cos 68 x
=
1
9.6
x = 9.6 cos 68 = 3.596  3.6 cm
cos 68 =
To find 9.6 cos 68 on a calculator, press
9.6 × COS 68 =
or
9.6 × 68 COS =
Example 3
Find A of ABC when AB = 44, AC = 29 and C = 90 .
Solution
Sketch ABC and label the given parts (Figure 19). The hypotenuse
and adjacent side to A are given, so use the cosine ratio to find A.
29
= 0.659
44
A = 48.8
cos A =
Figure 19
To find A, with the calculator displaying its cosine of 0.659, press INV COS or 2ndF COS or
COS−1 , to display the angle 48.8°.
When the cosine of an angle is known, the angle is called the “inverse cosine” or “arc cosine”.
A common mnemonic (memory assisting device) to remember the sine, cosine,
and tangent ratios is the word SOHCAHTOA (pronounced “sew caw toa”),
which is shorthand for the ratios
Opposite
Adjacent
Opposite
Sine =
, Cosine =
, Tangent =
Hypotenuse
Hypotenuse
Adjacent
Now complete Exercise 5.6 and check your answers.
27
Exercise 5.6
1.
Use a calculator to find the following pairs of sine and cosine values.
Round answers to four decimal places.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
sin A
sin 1
sin 5
sin 10
sin 22.5
sin 30
sin 45
sin 60
sin 67.5
sin 80
sin 85
sin 89
cos B
cos 89
cos 85
cos 80
cos 67.5
cos 60
cos 45
cos 30
cos 22.5
cos 10
cos 5
cos 1
Study both columns of answers. Notice that for each pair of values, A + B = 90
(“complementary” angles), and sin A = cos B.
For any pair of complementary angles, the sine of one angle equals the cosine of the other,
or in other words, sin A = cos (90 − A).
2.
From the triangle shown, find the following to one decimal place.
a.
b.
c.
d.
e.
sin 37
cos 53
sin 53
cos 37
suppose that sin 20 = 0.342. What is cos 70?
37
o
5
4
53
o
3
3.
Find side x in each triangle. Round your answers to one decimal place. For each of the
following triangles, ask yourself the following three questions:
1. Name side x. Is it the hypotenuse (H), the opposite side (O), or the adjacent side (A)?
2. Name the given side. Is it the hypotenuse, the opposite side, or the adjacent side?
3. Using SOHCAHTOA, which ratio should be used? sin, cos, or tan?
28
a. Name side x. Is it the hypotenuse,
opposite, or adjacent to 30?
Name the given side 40. Is it
the H, O or A side?
Which ratio? sin, cos or tan?
b. Name side x. H, O or A?
Name side 13. H, O or A?
Use sin, cos or tan?
40
13
X
o
25
o
30
X
Ask and answer the above three questions for each triangle below.
d.
c.
9
60
1
4
2
o
X
X
o
7
0
e.
f.
o
17
X
X
91
6.5
o
53
g.
h.
15.9
o
7
2
.5
X
o
45
3
.6
X
4.
In ABC C = 90, B = 24 and AC = 75 cm. Find AB .
5.
In ABC C = 90, A = 69 and AB = 24.7 cm. Find BC .
29
6.
Find x in each of the following. Round your answers to the nearest degree.
(use SOHCAHTOA)
b.
a.
17.3
X
20
15
30
X
c.
d.
305
X
5.1
3.6
462
X
e.
f.
95.3
X
4.85
83.4
X
6.02
7.
In ABC C = 90, AC = 14 and AB = 19. Find A and B to the nearest tenth of a
degree.
8.
A 10 m long ramp rises to a point 1 m
above the ground. Find the angle of
inclination of the ramp.
1
0
m
1
m
X
30
9.
An eagle is sitting on a tree 14 m above the ground. The eagle sights a mouse on the ground
at an angle of 58 with the tree. How far is the eagle from the mouse?
o
58
14 m
10.
An airplane travels at 450 km/hr in a
N50E direction. How far east has
the plane travelled after 1 hour?
N
450 km
50o
E
W
S
Answer key at the end of this unit.
31
5.7
Solving triangles
To “solve a triangle” means to determine all the unknown sides and angles. Before this is possible,
you have to know three parts of the triangle. One part must be a side..
Example 1
B
Solve ABC where C = 90, A = 53 and c = 18.
Solution
Sketch ABC and label the known parts (Figure 20), then find
B.
A + B = 90
53 + B = 90
B = 90 - 53
B = 37
c = 18
a
53
C
o
b
A
Figure 20
Find side a. Side a is opposite 53 and c = 18 is the hypotenuse. Using the sine ratio,
a
18
18 sin 53 = a
a  14.4
sin 53 =
There are several ways of determining side b. The safest method, which involves only given
information, is to use the cosine ratio, since side b is adjacent to 53.
b
18
18 cos 53 = b
b  10.8
cos 53 =
Both the Pythagorean Theorem and the tangent ratio require calculated answers to find side b. Even
though they are valid methods, they are not the most reliable. You could use a formula such as:
a
cos 37 =
to find side a.
18
This requires that you involve the calculated value of 37. If 37 is a wrong answer, then side a will
also be a wrong answer. It is always more reliable to involve given information in a calculation
whenever possible. Also note that if you round a value that you use in a subsequent calculation, the
next value that you find will also be an approximation.
32
Example 2
Solve the triangle in Figure 21.
Solution
Find the hypotenuse, RT as follows:
RT = 10 2 + 152
Figure 21
2
RT = 100 + 225 = 325
RT =
325  18.0
Find R as follows; side 15 is opposite R, side 10 is adjacent to R, so use the tangent ratio,
15
= 1.5
10
R  56.3
tan R =
You can use the tangent ratio to find T:
10
15
T  33.7
tan T =
Although it is not as reliable, it is also just as easy to use:
T + R = 90
T + 56.3 = 90
T = 90 - 56.3 = 33.7
Now complete Exercise 5.7 and check your answers.
33
Exercise 5.7
1.
Solve the following triangles. Find the angles to the nearest degree and the sides to one
decimal place.
a.
b.
A
a = 25
C
B
o
40
c
b
b
c = 28
A
C
B
a=5
c.
d.
b
A
B
C
c
a
c = 16
o
65
a = 35
B
A
e.
A
C
b = 40
f.
A
b
23o
C
b = 7.6
c
c
a = 60
o
42
C
B
B
a
2.
Solve ABC where C = 90, A = 16 and b = 92.
3.
Solve ABC where C = 90, a = 12 and c = 13.
4.
Solve ABC where C = 90, a = 1 and b = 1.
34
5.
Solve ABC where C = 90, A = 40 and B = 50.
6.
In ABC, note that ACB = 90, A = 60 and altitude CD = 12. Find the following:
a. B
b. AC
c. BC
d. AB
7.
In ABC, ACB = 90, A = 65 and AB
= 20. Find altitude CD .
8.
An airplane flies 300 km at a
heading of 80. How far north and
how far east has it travelled?
Answer key at the end of this unit.
Now complete the Practice Test and check your answers.
35
Practice Test
1.
Use a calculator to find the following. Round your answers to four decimal places.
a.
b.
c.
d.
2.
sin 36
cos 9.2
tan 87
sin 0.49
Use a calculator to find A in each of the following. Round your answers to the nearest
degree.
a. cos A = 0.9889
b. tan A = 1.5901
6
c. sin A =
11
d. tan A = 3
3.
Find side x or angle x in each of the following. Round your answers to one decimal place.
Remember SOHCAHTOA.
36
4.
Solve ABC where C = 90, B = 40 and a = 88.
5.
Solve ABC where C = 90, a = 24 and c = 30.
6.
Find the altitude CD of ABC if A = 25 and AB = 1.2. Note that ABC is a right
triangle.
C
B
A
D
7.
Engineers measure the right-angled distances beside a mountain as shown. They propose to
build a tunnel from A to C through the mountain. What is the length of the tunnel?
8.
Grace wants to know how far it is to the other side of the river. She makes the
measurements as shown. What is the distance from A to B across the river?
A
o
35
C
20 m
B
Answer key at the end of this unit.
37
Problem Set A
Read each question carefully, then draw a diagram. When you have finished the diagram, read the
question again to make sure you have interpreted it properly, then work out the solution.
1.
Pat wants to know the height of a tree in her yard. She observes that from a distance of 20
m from the tree, the angle of elevation to the top of the tree is 32. What is the height of the
tree to the nearest tenth of a metre?
2.
From a lighthouse 16 m above sea level, the angle of depression to a small boat is 12. How
far from the foot of the lighthouse is the boat to the nearest metre?
3.
A kite is 40 m high when 210 m of string is used. What angle does the kite string make with
the horizontal to the nearest degree?
4.
Marianne wants to know how far it is across a river. She notices a tree at point X straight
across from point Y. She walks 10 m along the river bank to point Z and observes that the
angle to the tree is 65. What is the distance across the river from point X to point Y to the
nearest hundredth of a metre?
X
River
65o
Y
5.
Z
10 m
A plot of land has the shape of a right triangle. The longest side is 37 m and lies at an angle
of 53 to the shortest side. Find the area of the plot to the nearest square metre.
38
6.
A painter is using a ladder to paint a high factory wall. The ladder is 5 m long and for
safety, must never be used at an angle to the vertical of less than 15 or more than 40. If he
can reach 1 m above the top of the ladder, what is the maximum height to the nearest tenth
of a metre that he can paint? (Hint: what angle would give the painter the greatest height?)
7.
Jack is building a garden shed of his own design. It has a simple sloping roof. The two
walls on which the roof will rest are 3.2 m apart and one wall is 0.5 m higher than the other.
Allowing 0.25 m for overhang at either end, how long (to the nearest hundredth) will the
roof beams have to be? What will the slope of the roof be to the nearest degree with respect
to the horizontal?
Slopingroof
Shed
8.
Joan is welding a piece of modern sculpture. Part of the design includes an A-frame
structure. Joan wants the two thin bars that make up the sides of the frame to form an angle
of 54 at the top and she wants the frame to be 2.2 m high. How long will each bar have to
be to the nearest thousandth? (Hint: you need a right triangle to use a trigonometric ratio.)
9.
A new ski-lift is being built at the slopes. The base of the lift is at an elevation of
2 500
m, but the elevation of the top station is not accurately known. A survey of the site shows
the base and the top station are 2 450 m apart in horizontal distance and a line of sight to the
top station angles up at 38. Find the length of steel cable (to the nearest 10 m) that will be
needed for the endless loop on which the chairs will hang. Allow for an additional 5% of
the total length for sags, joining, etc.
10.
A mathematical spider spinning a web across an alleyway between two buildings manages
to stretch a length of silken thread across from the top of a 2 m high doorway to a point on
the wall of the opposite building 1 m above the ground. The spider just happened to make
the angle of the thread exactly 25 to the horizontal. How far apart are the buildings to the
nearest centimetre?
11.
The roof of a small pup tent is made of a rectangular piece of material. If the tent is to be
2.2 m long, the roof sloping up at 48 to the horizontal and with the poles 1.4 m high, how
many square metres of material will be needed to make the tent roof? (nearest hundredth)
39
Problem Set B
1.
Given that the property line is halfway between the two houses on the plan below, what is
the distance between the two houses to the nearest tenth of a metre?
Property line
House A
House B
19.4 m
72o
Survey post
Road
2.
How far is the corner of house B from the survey post to the nearest tenth of a metre?
3.
A carpenter is instructed to cut a right-angled wooden wedge 25 cm long in the base with an
angle of 12. How long will the sloping surface of the wedge be to the nearest millimetre?
4.
From a ladder, Wayne looks at a building 35 m away. He notes that the angle of elevation to
the top of the building is 19 and the angle of depression to the bottom of the building is 7.
How high is the building?
5.
A slide in the Water Park is 9 m high. If the actual length of the slide is 14 m, what angle
does the slide make with the horizontal
40
Appendix A: Table of Trigonometric Ratios
Angle
sin
cos
tan
Angle
sin
cos
tan
0
1
2
3
4
5
0.000
0.017
0.035
0.052
0.070
0.087
1.000
1.000
0.999
0.999
0.998
0.996
0.000
0.017
0.035
0.052
0.070
0.087
45
46
47
48
49
50
0.707
0.719
0.731
0.743
0.755
0.766
0.707
0.695
0.682
0.669
0.656
0.643
1.000
1.036
1.072
1.111
1.150
1.192
6
7
8
9
10
0.105
0.122
0.139
0.156
0.174
0.995
0.993
0.990
0.988
0.985
0.105
0.123
0.141
0.158
0.176
51
52
53
54
55
0.777
0.788
0.799
0.809
0.819
0.629
0.616
0.602
0.588
0.574
1.235
1.280
1.327
1.376
1.428
11
12
13
14
15
0.191
0.208
0.225
0.242
0.259
0.982
0.978
0.974
0.970
0.966
0.194
0.213
0.231
0.249
0.268
56
57
58
59
60
0.829
0.839
0.848
0.857
0.866
0.559
0.545
0.535
0.515
0.500
1.483
1.540
1.600
1.664
1.732
16
17
18
19
20
0.276
0.292
0.309
0.326
0.342
0.961
0.956
0.951
0.946
0.940
0.287
0.306
0.325
0.344
0.364
61
62
63
64
65
0.875
0.883
0.891
0.899
0.906
0.485
0.469
0.454
0.438
0.423
1.804
1.881
1.963
2.050
2.145
21
22
23
24
25
0.358
0.375
0.391
0.407
0.423
0.934
0.927
0.921
0.914
0.906
0.384
0.404
0.424
0.445
0.466
66
67
68
69
70
0.914
0.921
0.927
0.934
0.940
0.407
0.391
0.375
0.358
0.342
2.246
2.356
2.475
2.605
2.747
26
27
28
29
30
0.438
0.454
0.469
0.485
0.500
0.899
0.891
0.883
0.875
0.866
0.488
0.510
0.532
0.554
0.577
71
72
73
74
75
0.946
0.951
0.956
0.961
0.966
0.326
0.309
0.292
0.276
0.259
2.904
3.078
3.271
3.487
3.732
31
32
33
34
35
0.515
0.530
0.545
0.559
0.574
0.857
0.848
0.839
0.829
0.819
0.601
0.625
0.649
0.675
0.700
76
77
78
79
80
0.970
0.974
0.978
0.982
0.985
0.242
0.225
0.208
0.191
0.174
4.011
4.332
4.705
5.145
5.671
36
37
38
39
40
0.588
0.602
0.616
0.629
0.643
0.809
0.790
0.788
0.777
0.766
0.727
0.754
0.781
0.810
0.839
81
82
83
84
85
0.988
0.990
0.993
0.995
0.996
0.156
0.139
0.122
0.105
0.087
6.314
7.115
8.144
9.514
11.430
41
42
43
44
45
0.656
0.669
0.682
0.695
0.707
0.755
0.743
0.731
0.719
0.707
0.869
0.900
0.933
0.966
1.000
86
87
88
89
90
0.998
0.999
0.999
1.000
1.000
0.070
0.052
0.035
0.017
0.000
14.301
19.081
28.636
57.290
NOTE: values are correct to the nearest 1/1000th
41
Appendix B: Glossary
Acute angle
Any angle whose measure is greater than 0 and less than 90. A right triangle always has
two acute angles and one right angle. In Figure 22, A and B are acute angles.
B
Cosine of an angle
For a specified acute angle of a right triangle, the cosine of the angle
is equal to the ratio of the length of the adjacent leg to the length of
the hypotenuse. For the triangle in Figure 22:
cos A =
b
c
c
a
A
b
Figure 22
Hypotenuse
The longest side of a triangle, opposite the right angle. In Figure 22, side c is the hypotenuse.
Leg
Either of the two sides of a right triangle that are opposite the two acute angles. In Figure 22,
sides b and a are the legs.
Pythagorean Theorem
The Pythagorean Theorem states that the square of the length of the hypotenuse is equal to the
sum of the squares of the length of the legs. For the triangle in Figure 22:
c2 = a2 + b2
Right triangle
A triangle that has a 90 angle. Figure 22 is a right triangle.
Sine of an angle
For a specified angle of a right angle triangle, the sine of the angle is equal to the ratio of the
length of the opposite leg to the length of the hypotenuse. For the triangle in Figure 22:
a
sin A =
c
Tangent of an angle
For a specified angle in a right angle triangle, the tangent of an angle is equal to the ratio of
the length of the opposite side to the length of the adjacent side. For the triangle in Figure 22:
a
tan A =
b
42
C
Answers
Exercise 5.1 - page 8
1. T
7. T
2. T
8. T
3. T
9. T
Exercise 5.2 - pages 10-11
4. T
10. T
1. a. c or AB
2. a. PQ or r
b. b or AC
b. PR or q
c. b or AC
c. PR or q
f. PQ or r
3. a. 60
4. a. large or L
g. 70
b. 80
c. 15
b. middle or M
5. F
6. T
d. a or BC
d. QR or p
e. 50
e. QR or p
d. 90
e. 36
c. short or S d. yes
f. 45
Exercise 5.3 - pages 14-16
1. a. C
b. O
c. B
d. Y
2. a. c
b. AB
c. PR
d. e
3. a. 18.9
b. 34.0
c. 12.3 d. 74.4
e. 21.2 f. 1.1
g. 13.2
h. 8.4
4. 65.6 cm
5. 3.8 m
6. 11.5 km
7. Yes, the distance between opposite corners is more than 2.4 m.
Exercise 5.4 - pages 19-20
1. calculator answers to 6 decimal places: tan 10 = 0.176327 tan 20 = 0.363970,
tan 45 = 1.000000
tan 50 = 1.191754
tan 75 = 3.732051
2. a. 0.0000
b. 0.0175
c. 0.0875
d. 0.2679
e. 0.4040
f. 0.6104
g. 1.0000
h. 1.2314
i. 2.0503
j. 0.0009
k. 14.3007
l. 57.2900
3. a. tan 25 = 0.5
b. tan 65 = 2.1
4. tan 88 = 28.636
tan 89 = 57.290
tan 89.9 = 572.957
tan 89.99 = 5729.578 tan 90 undefined (your calculator should display ERROR)
5. tan 90 = undefined. Since it is impossible to draw a triangle with two 90 angles, tan
90 cannot be calculated.
43
Exercise 5.5 - pages 23-25
1. a. tan 15 = x/10, x = 2.7
d. tan 40 = x/17.3, x = 14.5
g. tan 80 = 982/x, x = 173.2
2. a. BC = 12.6
3. a. 34.4
b. 54
4. A = 54.8
B = 35.2
5. B = 18
AC = 4.5
6. A = 40
AC = 88.2
7. 6.8
8. 27.5 m + 1.6 m = 29.1 m
9. 31
b.
e.
h.
b.
c.
tan 30 = 8/x, x = 13.9
c. tan 65 = x/19, x = 40.7
239
tan 53 = /x, x = 180.1 f. tan 39 = x/10.9, x = 8.8
tan 13 = x/50.7, x = 11.7
BC = 9.1
40.6
d. 69
AB = 14.7
AB = 115.1
Exercise 5.6 - pages 28-31
1. sin 1
0.0175
cos 89
0.0175
sin 5
0.0872
cos 85
0.0872
sin 10
0.1736
cos 80
0.1736
sin 22.5
0.3827
cos 67.5
0.3827
sin 30
0.5000
cos 60
0.5000
sin 45
0.7071
cos 45
0.7071
sin 60
0.8660
cos 30
0.8660
sin 67.5
0.9239
cos 22.5
0.9239
sin 80
0.9848
cos 10
0.9848
sin 85
0.9962
cos 5
0.9962
sin 89
0.9998
cos 1
0.9998
Notice that both answer columns are the same. Also notice that sin 1 = cos 89 and
that 1 + 89 = 90, sin 5 = cos 85 and 5 + 85 = 90 and so on down the columns.
2. a. sin 37 = 3/5 = 0.6
b. cos 53 = 3/5 = 0.6
c. sin 53 = 4/5 = 0.8
d. cos 37 = 4/5 = 0.8
Notice that sine 37 = cos 53 since 37 + 53 = 90. Also sin 53 = cos 37 because
53 + 37 = 90.
e. cos 70 = 0.342 since 20 + 70 = 90
3. a. sin 30 = x/40, x = 20
b. cos 25 = x/13, x = 11.8
c. tan 60 = x/9, x = 15.6
d. sin 70 = 142/x,x = 151.1
e. cos 17 = 91/x, x = 95.2
f. sin 53 = x/6.5, x = 5.2
15.9
3.6
g. cos 45 = /x, x = 22.5
h. sin 72.5 = /x, x = 3.8
4. 184.4 cm
5. 23.1 cm
6. a. sin x = 15/20, x = 49
b. cos x = 17.3/30, x = 55 c. cos x= 3.6/5.1, x = 45
305
d. sin x = /462, x = 41
e. cos x = 4.85/6.02, x = 36 f. cos x = 83.4/9.5, x = 29
7. A = 42.5 and B = 47.5
8.
5.7
9. 26.4 m
10. 345 km
44
Exercise 5.7 - pages 34-35
1. a. B = 50, b = 6, c = 7.8
b. A = 63, B = 27, b = 12.6
c. A = 25, a = 6.8, b = 14.5
d. c = 53.2, A = 41, B = 49
e. A = 48, b = 54, c = 80.7
f. B = 67, a = 3.2, c = 8.3
2. B = 74, a = 26.4, c = 95.7
3. b = 5, A = 67, B = 23
4. c = 1.4, A = 45, B = 45
5. There are an infinite number of 40-50-90 triangles.
6. B = 30, AC = 13.9, BC = 24, AB = 27.7
7. BC = 20 sin 65 = 18.126 and B = 25, therefore CD = 7.66
8. 52.1 km north and 295.4 km east
Practice Test - pages 36-37
1. a. 0.5878
b. 0.9871
c. 19.0811
d. 0.0086
2. a. 9
b. 58
c. 33
d. 72
3. a. 34.2
b. 6
c. 39.3
d. 54.3
e. 51.2
f. 56.7
4. A = 50, b = 73.8, c = 114.9
5. b = 18, A = 53, B = 37
AC
CD
6. cos 25 =
so AC = 1.088. Now sin 25 =
and CD = 0.46
1.2
1.088
7. 2.8 km
8. 14 m
Problem Set A – pages 38-39
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.5 m
75 m
11
21.45 m
329 m2
5.8 m
roof beams are 3.73 m long, slope is 9
2.469 m
6530 m
214 cm
8.27 m2
Problem Set B – page 400
1.
2.
3.
4.
5.
28.2 m
24.1 m
256 mm
16.3 m
40
45
Notes